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I am new to combinatorics and struggling with the following quastion:

How many different functions $$f:\left\{0,1\right\}^{n}\rightarrow \left\{0,1\right\}^{m}$$ can be found? if: $$m \geq n$$

How I imagine it:

every $\left\{0,1\right\}^{n}$ is a series of $1$-th and $0$-th in the length of $n$, so for every element we can choose to change it to ($0$ or $1$) or to leave it without changing it.

So for every element we got $2$ options and we got $n$ elements so we got $2^{n}$ options correct?

Now, the moment we finish all the elements ($n$) for every new place we got $2$ options - $0$ or $1$ and we are left with $m-n$ places to fill so we got $2^{m-n}$ options, correct?

The final answer is : $$2^{m}\cdot2^{m-n}=2^{m}$$

Is this correct or I completely misunderstood some-thing?

misha312
  • 537

1 Answers1

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Let $A,B$ finite sets, the number of function $f:A \rightarrow B$ is $|B|^{|A|}$ (why?).

In this case the number of considered function is $(2^m)^{2^n}$

ictibones
  • 1,984